3.2.5 \(\int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\) [105]

Optimal. Leaf size=47 \[ -\frac {x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3} \]

[Out]

-1/2*x^2/b/arctanh(tanh(b*x+a))^2-x/b^2/arctanh(tanh(b*x+a))+ln(arctanh(tanh(b*x+a)))/b^3

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Rubi [A]
time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 29} \begin {gather*} \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x}{b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

-1/2*x^2/(b*ArcTanh[Tanh[a + b*x]]^2) - x/(b^2*ArcTanh[Tanh[a + b*x]]) + Log[ArcTanh[Tanh[a + b*x]]]/b^3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[b*(n/(a*(m + 1))), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {\int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{b}\\ &=-\frac {x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {\int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=-\frac {x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}\\ &=-\frac {x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 49, normalized size = 1.04 \begin {gather*} \frac {3-\frac {b^2 x^2}{\tanh ^{-1}(\tanh (a+b x))^2}-\frac {2 b x}{\tanh ^{-1}(\tanh (a+b x))}+2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(3 - (b^2*x^2)/ArcTanh[Tanh[a + b*x]]^2 - (2*b*x)/ArcTanh[Tanh[a + b*x]] + 2*Log[ArcTanh[Tanh[a + b*x]]])/(2*b
^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(135\) vs. \(2(45)=90\).
time = 0.11, size = 136, normalized size = 2.89

method	result	size
default	\(\frac {2 a}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )-2 b x -2 a}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {a^{2}}{2 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{b^{3}}\)	\(136\)
risch	\(-\frac {4 i \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) x -\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x +\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) x -2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2} x +\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3} x -\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x +\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3} x +4 i x \ln \left ({\mathrm e}^{b x +a}\right )+2 i b \,x^{2}\right )}{b^{2} \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )^{2}}+\frac {\ln \left (\ln \left ({\mathrm e}^{b x +a}\right )+\frac {i \pi \left (-\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}\right )}{4}\right )}{b^{3}}\)	\(816\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arctanh(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

2/b^3/arctanh(tanh(b*x+a))*a+2/b^3/arctanh(tanh(b*x+a))*(arctanh(tanh(b*x+a))-b*x-a)-1/2/b^3/arctanh(tanh(b*x+
a))^2*a^2-1/b^3/arctanh(tanh(b*x+a))^2*a*(arctanh(tanh(b*x+a))-b*x-a)-1/2/b^3/arctanh(tanh(b*x+a))^2*(arctanh(
tanh(b*x+a))-b*x-a)^2+ln(arctanh(tanh(b*x+a)))/b^3

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Maxima [A]
time = 0.85, size = 48, normalized size = 1.02 \begin {gather*} \frac {4 \, a b x + 3 \, a^{2}}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {\log \left (b x + a\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

1/2*(4*a*b*x + 3*a^2)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) + log(b*x + a)/b^3

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Fricas [A]
time = 0.38, size = 61, normalized size = 1.30 \begin {gather*} \frac {4 \, a b x + 3 \, a^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/2*(4*a*b*x + 3*a^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

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Sympy [A]
time = 39.21, size = 54, normalized size = 1.15 \begin {gather*} \begin {cases} - \frac {x^{2}}{2 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \operatorname {atanh}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/atanh(tanh(b*x+a))**3,x)

[Out]

Piecewise((-x**2/(2*b*atanh(tanh(a + b*x))**2) - x/(b**2*atanh(tanh(a + b*x))) + log(atanh(tanh(a + b*x)))/b**
3, Ne(b, 0)), (x**3/(3*atanh(tanh(a))**3), True))

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Giac [A]
time = 0.38, size = 37, normalized size = 0.79 \begin {gather*} \frac {\log \left ({\left | b x + a \right |}\right )}{b^{3}} + \frac {4 \, a x + \frac {3 \, a^{2}}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

log(abs(b*x + a))/b^3 + 1/2*(4*a*x + 3*a^2/b)/((b*x + a)^2*b^2)

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Mupad [B]
time = 1.03, size = 46, normalized size = 0.98 \begin {gather*} \frac {\ln \left (\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^3}-\frac {\frac {b^2\,x^2}{2}+b\,x\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{b^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/atanh(tanh(a + b*x))^3,x)

[Out]

log(atanh(tanh(a + b*x)))/b^3 - ((b^2*x^2)/2 + b*x*atanh(tanh(a + b*x)))/(b^3*atanh(tanh(a + b*x))^2)

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